Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 01(q0(x1))
11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q2(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
11(q1(1(x1))) → 11(q1(x1))
01(q1(x1)) → 11(x1)
11(q1(1(x1))) → 11(1(q1(x1)))

The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 01(q0(x1))
11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q2(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
11(q1(1(x1))) → 11(q1(x1))
01(q1(x1)) → 11(x1)
11(q1(1(x1))) → 11(1(q1(x1)))

The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q2(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q1(1(x1))) → 11(q1(x1))
01(q1(x1)) → 11(x1)
11(q1(1(x1))) → 11(1(q1(x1)))

The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q2(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
01(q1(x1)) → 11(x1)
The remaining pairs can at least be oriented weakly.

11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(1(x1))) → 11(q1(x1))
11(q1(1(x1))) → 11(1(q1(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(q1(x1)) = 1/2 + x_1   
POL(11(x1)) = (1/4)x_1   
POL(1(x1)) = x_1   
POL(01(x1)) = (1/4)x_1   
POL(q0(x1)) = 3/4 + x_1   
POL(q2(x1)) = 3/4 + x_1   
POL(0(x1)) = 1/4 + x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q2(x1)) → q2(1(x1))
0(q1(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(1(x1))) → 11(q1(x1))
11(q1(1(x1))) → 11(1(q1(x1)))

The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

11(q1(1(x1))) → 11(q1(x1))

The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


11(q1(1(x1))) → 11(q1(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(q1(x1)) = (2)x_1   
POL(11(x1)) = (2)x_1   
POL(1(x1)) = 1/4 + (11/4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

1(q0(1(x1))) → 0(1(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
0(q1(x1)) → q2(1(x1))
1(q2(x1)) → q2(1(x1))
0(q2(x1)) → 0(q0(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.